For a multinomial distribution
$P = ( n! / (\prod_{i=1}^k {n_i!}) ) * ( \prod_{i=1}^k {{p_i}^{n_i}})$
If you sum the resulting multinomial distribution for every possible unique frequency for k=2 for any integer n that is n > 1 (this being equivalent to the binomial distribution) this will equal 1.00. The same is true for k=3
But for integers k>3 this does not seem to hold. It will always be less than 1.00. Try it yourself for probabilities <0.25,0.25,0.25,0.25> and n=4.
You’ll get $0.09375 + (0.046875*12) + (0.01563*12) + (0.00391*4) = 0.85951$
<1,1,1,1> -> 1 unique permutation adding up to 4
<0,1,1,2> -> 12 unique permutations adding up to 4
<0,0,1,3> -> 12 unique permutations adding up to 4
<0,0,0,4> -> 4 unique permutations adding up to 4
I multiplied by 12,12,and 4 because they have the same multinomial distribution as you’ll see when you try it.
What’s going on here? I’ve made a mistake right?
—-please excuse the informalities I’m not a mathematician
They have to sum to one since
$$\sum_{n_i\in \{0,...,n\}:n_1+...+n_k=n}{n\choose n_1,n_2,...,n_k} p_1^{n_1}\cdot...\cdot p_k^{n_k}=(p_1+...+p_k)^n=1.$$
In your example, using your notation, you forgot the case $<2,2,0,0>$ for which there are 6 permutations. Thus, the LHS becomes
$$0.25^4 \left({4\choose 1,1,1,1}+12{4\choose 0,1,1,2}+6{4\choose 2,2,0,0}+12{4\choose 0,0,1,3}+4{4\choose 0,0,0,4}\right)=1$$